Question

Calculate the free energy change at $298K$ for the reaction :

$B{r}_2\left(l\right)+C{l}_2\left(g\right)\to 2BrCl\left(g\right)$. For the reaction $\Delta H^o=29.3kJ$ & the entropies of

$B{r}_2\left(l\right),C{l}_2\left(g\right)$ & BrCl(g) at the $298K$ are $152.3,223.0,239.7$ $Jmo{l}^1{K}^1$ respectively

• A. $1721.8J$
• B. $1721.8kJ$
• C. $17218J$
• D. None of these

Answer 1

The correct option is A $1721.8J$

$B{r}_2\left(l\right)+C{l}_2\left(g\right)\to 2BrCl\left(g\right),\Delta H^o=29.3kJ$

$S_{Br}=152.3S_{C{l}_2\left(g\right)}=223.0$

$S_{BrCl\left(g\right)}=239.7Jmo{l}^1{K}^1$

The entropy change for the reaction is

$\Delta S_R=2S_{BrCl\left(g\right)}-[S_{Br}+S_{C{l}_2\left(g\right)}]=2\times 239.7-[223.0+152.3]=104.4Jmo{l}^1{K}^1$

The relationship between the free energy change, the enthalpy change and the entropy change is as shown.

${\Delta }_{r}G=\Delta H-T\Delta S$

Substitute values in the above expression.

${\Delta }_{r}G=29300298\times 104.4=1721.8J$

Hence, the free energy change for the reaction is $1721.8J$